- Email: [email protected]
Corrigendum to “Variations on a theorem by Alan Camina on conjugacy class sizes” [J. Algebra 296 (2006) 253–266]
Journal of Algebra 320 (2008) 4317–4319
Contents lists available at ScienceDirect
Journal of Algebra www.elsevier.com/locate/jalgebra
Corrigendum
Corrigendum to “Variations on a theorem by Alan Camina on conjugacy class sizes” [J. Algebra 296 (2006) 253–266] ✩ Antonio Beltrán a,∗ , María José Felipe b a b
Departamento de Matemáticas, Universidad Jaume I, 12071 Castellón, Spain Departamento de Matemática Aplicada and IMPA-UPV, Universidad Politécnica de Valencia, 46022 Valencia, Spain
a r t i c l e
i n f o
Article history: Received 7 August 2008 Available online 28 October 2008 Keywords: Group theory Conjugacy class sizes
The authors have noticed that in several steps of the proofs of Theorems A and B in the published version minor cases appear which are not considered, and however these require some non-trivial arguments which are not reflected in the proofs of those steps. All of them make reference to the following situation. Suppose that all conjugacy class sizes or that all conjugacy class sizes of p elements for some prime p of some group may be equal to two numbers, say 1 or n. Then we use the main theorem of [2]. But the case in which all of them are exactly equal to 1, although it may seem easier, actually remains unsolved. These are precisely the cases we are completing here. Proof of Theorem A. The published proof of Theorem A requires the following change in Step 1 so that we can correct the cases pointed above. We denote by π the set of primes dividing n. Step 1. G does not posses abelian Hall π -subgroups and if G has a normal Hall π -subgroup then G satisfies the conclusions of the theorem. The same properties hold for Hall π -subgroups. Suppose that G has an abelian Hall π -subgroup H and thus any element of H has index 1 or m in G. If H is a central subgroup, then n = 1 and this is not possible, so we can take y ∈ H such ✩ This work was partially supported by grant MTM2007-68010-C03-03 and the first author is also supported by grant Fundació Caixa-Castelló P1-1A2006-06. DOI of original article: 10.1016/j.jalgebra.2005.06.031. Corresponding author. E-mail addresses: [email protected] (A. Beltrán), [email protected] (M.J. Felipe).
*
0021-8693/$ – see front matter © 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.jalgebra.2008.10.003
4318
A. Beltrán, M.J. Felipe / Journal of Algebra 320 (2008) 4317–4319
that | y G | = m. Notice that by the primary decomposition we can assume without loss of generality that y is a q-element for some prime q ∈ π . Now, if x is a q -element of C G ( y ), then as C G (xy ) = C G (x) ∩ C G ( y ) ⊆ C G ( y ), we have that x has necessarily index 1 or n in C G ( y ). We are going to see that C G ( y ) = H × K y , where K y is a π -subgroup. We distinguish two cases. If one of all these indexes is n then, as C G ( y ) is a solvable group, we can apply Theorem 4 of [1] to obtain that n = qa r b , with r ∈ π a prime distinct from q, and C G ( y ) = Q R × T , where Q and R are Sylow q and r subgroups of G, respectively, and T is abelian. Since the Hall π -subgroups are abelian, we conclude that C G ( y ) can be written as C G ( y ) = H × K y , where K y is a π -subgroup. In the other case, that is, when C G (xy ) = C G ( y ) for every q -element x ∈ C G ( y ), the same result certainly follows. Let us take an element z of index mn and factor z = zπ zπ . If zπ is non-central, then it has index m and by the above paragraph we know that C G ( zπ ) = H × K zπ , where H is an abelian Hall π -subgroup of G. Since, zπ ∈ C G ( zπ ), we deduce that H ⊆ C G ( z) which leads to a contradiction. Thus, zπ can be assumed to be central in G and z can be replaced by zπ . Now, we see that C G ( z) = K z × V z , with V z an abelian π -subgroup and K z a π -subgroup. This follows since if t is any π -element of C G ( z), then C G (tz) = C G (t ) ∩ C G ( z) ⊆ C G ( z) and since mn is the largest class size of G, then C G (tz) = C G ( z), so C G ( z) ⊆ C G (t ), that is, t ∈ Z(C G (x)) and the assertion is proved. Suppose now that there exists some non-central t ∈ V z . We know that t has index m and, by the first paragraph, C G (t ) = H × K t with H an abelian Hall π -subgroup of G and K t a π -subgroup, so z ∈ K t and it follows that H ⊆ C G ( z) which is a contradiction. Therefore, we have V z = Z(G )π and in particular |G : Z(G )|π = n. We take y a π -element of index m and we consider C G ( y ) = H × K y as in the first paragraph. Notice that if K y ⊆ Z(G ) this would imply |G : Z(G )| = mn which yields a contradiction with the existence of elements of index mn. Thus, if we take some non-central w ∈ K y , then H ⊆ C G ( w ) ∩ C G ( y) = C G ( w y) ⊆ C G ( y) and this implies that C G ( w ) = C G ( y ). On the other hand, let K be a Hall π -subgroup of G such that K y ⊆ K . We prove that Z( K ) = Z(G )π . If α ∈ Z( K ), then α ∈ C G ( w ) = C G ( y ) and accordingly α ∈ K y . Thus G = H K = C G (α ) and α ∈ Z(G ), as wanted. Now, if t has index n, by using the {π , π }decomposition of t, we can suppose that t is a π -element, which notice that must lie in the center of some Hall π -subgroup of G. This contradiction finishes the proof of the first part of the step. Suppose now that G has a normal Hall
π -subgroup H . For every x ∈ H we have
|G : H | H : C H (x) = G : C G (x)C G (x) : C H (x). If |xG | = 1 or m, then H ⊆ C G (x) and thus |x H | = 1. If |xG | = n or mn, then the above equality along with the fact that |x H | divides |xG | imply that |x H | = n. Therefore, any conjugacy class of π -elements in H has size 1 or n. If all of them are equal to 1, then H is abelian but we have just proved that this cannot happen. Thus, we can apply Theorem 3 and we get n = qb for some prime q. Then we can apply Theorem B to obtain that G is nilpotent and m = pa , so the theorem is proved. The same arguments are valid for π instead of π , since we can argue symmetrically. Step 2. We may assume that there are no of index m.
π -elements of index n and that there are no π -elements
Suppose that x is a π -element of index n. By considering the primary decomposition of x we can assume without loss that x is a p-element for some prime p ∈ π . Now if y is a p -element of C G (x), then C G (xy ) = C G (x) ∩ C G ( y ) ⊆ C G (x) and this forces |C G (x) : C G (x) ∩ C G ( y )| = 1 or m. When all these indexes are equal to 1, it follows that C G (x) has an abelian π -complement and thus G has abelian π -complements, which is a contradiction by Step 1. We can apply then Theorem 4 and we obtain that m = p c qb , but as (n, m) = 1, then m = qb and G would be nilpotent by applying Theorem B. In this case we have necessarily n = pa for some a > 0. The second assertion is also true since we can argue symmetrically with m and n.
A. Beltrán, M.J. Felipe / Journal of Algebra 320 (2008) 4317–4319
4319
Steps 3 and 4 of the proof remain unchanged. Step 5. We may assume that n = qb r c for some distinct primes q and r. As a consequence of Step 2, we may choose a π -element, say x, of index m. It is enough to consider the decomposition of any element of index m as a product of a π -element by a π -element. In addition, if we consider the primary decomposition of x as a product of elements of prime power order, we can assume without loss that x is a q-element for some prime q ∈ π . Now if we take a q element w ∈ C G (x), we have C G ( wx) = C G ( w )∩ C G (x) and |C G (x) : C G ( w )∩ C G (x)| must be 1 or n. This proves that any q -element of C G (x) has index 1 or n in C G (x). Assume first that all these elements have index 1, and thus C G (x) can be written as C G (x) = Q × T , with Q a Sylow q-subgroup of C G (x) and T abelian. Now choose H a Hall π -subgroup of G with H ⊆ C G (x) and notice that H = Q × S with S abelian. For any h ∈ H we have S ⊆ C G (h), so |h G | must be a π ∪ {q}-number. This implies that either n is a q-number and the proof finishes by Theorem B, or every element of H has index 1 or n. By Lemma 1(a), it follows that H is abelian but this cannot happen by Step 1. Therefore, 1 and n are the indexes of q -elements of C G (x) and we can apply Theorem 4 to conclude that n = qb r c , with b, c 0, as wanted. Moreover, we can assume b, c > 0 by Theorem B. Thus the step is proved. The rest of the proof of Theorem A remains unchanged but now, once Step 7 is shown, we have a final contradiction with Step 1, so the proof of Theorem A finishes. Therefore, Step 8 can be eliminated. Proof of Theorem B. Theorem B is a special case of Theorem A and it is enough to add at the beginning of the proof, in the statement of Step 1, the following property: we can assume that G has no abelian p-complements. Its proof is analogous to the proof of Step 1 of Theorem A described above, by replacing the set π by { p } . The following steps stay invariable until Step 5, which has the same statement and now can be easily proved in a similar way to the proof Step 5 of Theorem A using the fact that G has no abelian p-complements. The remainder of the published proof of Theorem B stays equal except Step 9.1, which now can be eliminated as a consequence of Step 1 and Step 8.1. References [1] A. Beltrán, M.J. Felipe, Variations on a theorem of Alan Camina on conjugacy class sizes, J. Algebra 296 (2006) 253–266. [2] A. Beltrán, M.J. Felipe, Finite groups with two p-regular conjugacy class lengths, Bull. Austral. Math. Soc. 67 (2003) 163–169.
Contents lists available at ScienceDirect
Journal of Algebra www.elsevier.com/locate/jalgebra
Corrigendum
Corrigendum to “Variations on a theorem by Alan Camina on conjugacy class sizes” [J. Algebra 296 (2006) 253–266] ✩ Antonio Beltrán a,∗ , María José Felipe b a b
Departamento de Matemáticas, Universidad Jaume I, 12071 Castellón, Spain Departamento de Matemática Aplicada and IMPA-UPV, Universidad Politécnica de Valencia, 46022 Valencia, Spain
a r t i c l e
i n f o
Article history: Received 7 August 2008 Available online 28 October 2008 Keywords: Group theory Conjugacy class sizes
The authors have noticed that in several steps of the proofs of Theorems A and B in the published version minor cases appear which are not considered, and however these require some non-trivial arguments which are not reflected in the proofs of those steps. All of them make reference to the following situation. Suppose that all conjugacy class sizes or that all conjugacy class sizes of p elements for some prime p of some group may be equal to two numbers, say 1 or n. Then we use the main theorem of [2]. But the case in which all of them are exactly equal to 1, although it may seem easier, actually remains unsolved. These are precisely the cases we are completing here. Proof of Theorem A. The published proof of Theorem A requires the following change in Step 1 so that we can correct the cases pointed above. We denote by π the set of primes dividing n. Step 1. G does not posses abelian Hall π -subgroups and if G has a normal Hall π -subgroup then G satisfies the conclusions of the theorem. The same properties hold for Hall π -subgroups. Suppose that G has an abelian Hall π -subgroup H and thus any element of H has index 1 or m in G. If H is a central subgroup, then n = 1 and this is not possible, so we can take y ∈ H such ✩ This work was partially supported by grant MTM2007-68010-C03-03 and the first author is also supported by grant Fundació Caixa-Castelló P1-1A2006-06. DOI of original article: 10.1016/j.jalgebra.2005.06.031. Corresponding author. E-mail addresses: [email protected] (A. Beltrán), [email protected] (M.J. Felipe).
*
0021-8693/$ – see front matter © 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.jalgebra.2008.10.003
4318
A. Beltrán, M.J. Felipe / Journal of Algebra 320 (2008) 4317–4319
that | y G | = m. Notice that by the primary decomposition we can assume without loss of generality that y is a q-element for some prime q ∈ π . Now, if x is a q -element of C G ( y ), then as C G (xy ) = C G (x) ∩ C G ( y ) ⊆ C G ( y ), we have that x has necessarily index 1 or n in C G ( y ). We are going to see that C G ( y ) = H × K y , where K y is a π -subgroup. We distinguish two cases. If one of all these indexes is n then, as C G ( y ) is a solvable group, we can apply Theorem 4 of [1] to obtain that n = qa r b , with r ∈ π a prime distinct from q, and C G ( y ) = Q R × T , where Q and R are Sylow q and r subgroups of G, respectively, and T is abelian. Since the Hall π -subgroups are abelian, we conclude that C G ( y ) can be written as C G ( y ) = H × K y , where K y is a π -subgroup. In the other case, that is, when C G (xy ) = C G ( y ) for every q -element x ∈ C G ( y ), the same result certainly follows. Let us take an element z of index mn and factor z = zπ zπ . If zπ is non-central, then it has index m and by the above paragraph we know that C G ( zπ ) = H × K zπ , where H is an abelian Hall π -subgroup of G. Since, zπ ∈ C G ( zπ ), we deduce that H ⊆ C G ( z) which leads to a contradiction. Thus, zπ can be assumed to be central in G and z can be replaced by zπ . Now, we see that C G ( z) = K z × V z , with V z an abelian π -subgroup and K z a π -subgroup. This follows since if t is any π -element of C G ( z), then C G (tz) = C G (t ) ∩ C G ( z) ⊆ C G ( z) and since mn is the largest class size of G, then C G (tz) = C G ( z), so C G ( z) ⊆ C G (t ), that is, t ∈ Z(C G (x)) and the assertion is proved. Suppose now that there exists some non-central t ∈ V z . We know that t has index m and, by the first paragraph, C G (t ) = H × K t with H an abelian Hall π -subgroup of G and K t a π -subgroup, so z ∈ K t and it follows that H ⊆ C G ( z) which is a contradiction. Therefore, we have V z = Z(G )π and in particular |G : Z(G )|π = n. We take y a π -element of index m and we consider C G ( y ) = H × K y as in the first paragraph. Notice that if K y ⊆ Z(G ) this would imply |G : Z(G )| = mn which yields a contradiction with the existence of elements of index mn. Thus, if we take some non-central w ∈ K y , then H ⊆ C G ( w ) ∩ C G ( y) = C G ( w y) ⊆ C G ( y) and this implies that C G ( w ) = C G ( y ). On the other hand, let K be a Hall π -subgroup of G such that K y ⊆ K . We prove that Z( K ) = Z(G )π . If α ∈ Z( K ), then α ∈ C G ( w ) = C G ( y ) and accordingly α ∈ K y . Thus G = H K = C G (α ) and α ∈ Z(G ), as wanted. Now, if t has index n, by using the {π , π }decomposition of t, we can suppose that t is a π -element, which notice that must lie in the center of some Hall π -subgroup of G. This contradiction finishes the proof of the first part of the step. Suppose now that G has a normal Hall
π -subgroup H . For every x ∈ H we have
|G : H | H : C H (x) = G : C G (x)C G (x) : C H (x). If |xG | = 1 or m, then H ⊆ C G (x) and thus |x H | = 1. If |xG | = n or mn, then the above equality along with the fact that |x H | divides |xG | imply that |x H | = n. Therefore, any conjugacy class of π -elements in H has size 1 or n. If all of them are equal to 1, then H is abelian but we have just proved that this cannot happen. Thus, we can apply Theorem 3 and we get n = qb for some prime q. Then we can apply Theorem B to obtain that G is nilpotent and m = pa , so the theorem is proved. The same arguments are valid for π instead of π , since we can argue symmetrically. Step 2. We may assume that there are no of index m.
π -elements of index n and that there are no π -elements
Suppose that x is a π -element of index n. By considering the primary decomposition of x we can assume without loss that x is a p-element for some prime p ∈ π . Now if y is a p -element of C G (x), then C G (xy ) = C G (x) ∩ C G ( y ) ⊆ C G (x) and this forces |C G (x) : C G (x) ∩ C G ( y )| = 1 or m. When all these indexes are equal to 1, it follows that C G (x) has an abelian π -complement and thus G has abelian π -complements, which is a contradiction by Step 1. We can apply then Theorem 4 and we obtain that m = p c qb , but as (n, m) = 1, then m = qb and G would be nilpotent by applying Theorem B. In this case we have necessarily n = pa for some a > 0. The second assertion is also true since we can argue symmetrically with m and n.
A. Beltrán, M.J. Felipe / Journal of Algebra 320 (2008) 4317–4319
4319
Steps 3 and 4 of the proof remain unchanged. Step 5. We may assume that n = qb r c for some distinct primes q and r. As a consequence of Step 2, we may choose a π -element, say x, of index m. It is enough to consider the decomposition of any element of index m as a product of a π -element by a π -element. In addition, if we consider the primary decomposition of x as a product of elements of prime power order, we can assume without loss that x is a q-element for some prime q ∈ π . Now if we take a q element w ∈ C G (x), we have C G ( wx) = C G ( w )∩ C G (x) and |C G (x) : C G ( w )∩ C G (x)| must be 1 or n. This proves that any q -element of C G (x) has index 1 or n in C G (x). Assume first that all these elements have index 1, and thus C G (x) can be written as C G (x) = Q × T , with Q a Sylow q-subgroup of C G (x) and T abelian. Now choose H a Hall π -subgroup of G with H ⊆ C G (x) and notice that H = Q × S with S abelian. For any h ∈ H we have S ⊆ C G (h), so |h G | must be a π ∪ {q}-number. This implies that either n is a q-number and the proof finishes by Theorem B, or every element of H has index 1 or n. By Lemma 1(a), it follows that H is abelian but this cannot happen by Step 1. Therefore, 1 and n are the indexes of q -elements of C G (x) and we can apply Theorem 4 to conclude that n = qb r c , with b, c 0, as wanted. Moreover, we can assume b, c > 0 by Theorem B. Thus the step is proved. The rest of the proof of Theorem A remains unchanged but now, once Step 7 is shown, we have a final contradiction with Step 1, so the proof of Theorem A finishes. Therefore, Step 8 can be eliminated. Proof of Theorem B. Theorem B is a special case of Theorem A and it is enough to add at the beginning of the proof, in the statement of Step 1, the following property: we can assume that G has no abelian p-complements. Its proof is analogous to the proof of Step 1 of Theorem A described above, by replacing the set π by { p } . The following steps stay invariable until Step 5, which has the same statement and now can be easily proved in a similar way to the proof Step 5 of Theorem A using the fact that G has no abelian p-complements. The remainder of the published proof of Theorem B stays equal except Step 9.1, which now can be eliminated as a consequence of Step 1 and Step 8.1. References [1] A. Beltrán, M.J. Felipe, Variations on a theorem of Alan Camina on conjugacy class sizes, J. Algebra 296 (2006) 253–266. [2] A. Beltrán, M.J. Felipe, Finite groups with two p-regular conjugacy class lengths, Bull. Austral. Math. Soc. 67 (2003) 163–169.